Electrical Engineering - Circuit Theorems and Conversions - Discussion
Discussion Forum : Circuit Theorems and Conversions - General Questions (Q.No. 22)
22.
Referring to circuit given, if R1 is changed to a 68 
resistor, what will be the current through it?
resistor, what will be the current through it?
Discussion:
32 comments Page 1 of 4.
Ayman said:
1 decade ago
Why the current of the 0.2 amperes does not divide into two equal currents at the point between 68Ohm resistors, should not it?
Jyoti said:
1 decade ago
In each branch there should be equal currents.... means 0.1amp. Branch r2 already hve current source of .04 amp.
So current in ri is 0.1+.06 = 0.16amp
So current in ri is 0.1+.06 = 0.16amp
Sravanthi said:
1 decade ago
Why it should be 0.1 A in each branch ? please explain it clearly.
Vishal said:
1 decade ago
Jyoti can you explane how your gating 0.06 in last step?
L.Haque said:
1 decade ago
I1=.2A,when I1 active
I2=.04A,when I2 active
Final current=.2-.04=.16A(ans)
I2=.04A,when I2 active
Final current=.2-.04=.16A(ans)
Shiva said:
1 decade ago
Apply kcl at centre node
0.2-0.04=1.6
0.2-0.04=1.6
Siddu said:
1 decade ago
If resistance decreses current increases at node current flow 0.2 amps curent divide each branch then resistance is low curent flow high so 0.16
0.16 + 0.02 = 0.2 amps.
0.16 + 0.02 = 0.2 amps.
Ahmad jaradat said:
1 decade ago
v=.04*68=2.72 (fixed)
so
v at new 68 = 2.72 (parallel)
i=2.72\68 = .04 a
so
v at new 68 = 2.72 (parallel)
i=2.72\68 = .04 a
Drishti said:
1 decade ago
By kirchoff's current law.
I +0.2=0.04.
I=0.04-0.2 = -0.16A.
I +0.2=0.04.
I=0.04-0.2 = -0.16A.
Pradeep said:
1 decade ago
Apply kcl v1/68=2-.04.
v1/68 = 0.16.
v1 = 0.16*68.
v1 = i1*r1.
0.16*68 = i1*68.
i1 = 0.16A.
v1/68 = 0.16.
v1 = 0.16*68.
v1 = i1*r1.
0.16*68 = i1*68.
i1 = 0.16A.
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